%把所有的樣本都看作訓練集候選樣本，依次從中挑選樣本進訓練集，首先選擇歐氏距離最遠的兩個向量對進入訓練集，在接下來的迭代過程中擁有最大最小距離的待選樣本
%被選入訓練庫．以此類推，達到所要求的樣本數目。
%該方法優點是能保證訓練庫中樣本按照空間距離分布均勻。缺點是需要進行數據轉換和計算樣本兩兩空間距離，計算量大。

function?[mdminmax]?=?KS（XN）

%?Kennard-Stone?Algorithm?for?selection?of?samples
%?[mdminmax]?=?ks（XN）;
%
%?X?-->?Matrix?of?instrumental?responses
%?N?-->?Number?of?samples?to?be?selected?（minimum?of?2）
%
%?m?-->?Indexes?of?the?selected?samples
%
%?dminmax（1）?=?0;
%?dminmax（2）?=?Euclidean?distance?between?the?two?first?samples?selected?by?the?algorithm
%?dminmax（i）?=?Smallest?distance?between?the?i-th?selected?sample?and?the?previously?selected?ones?（i?>?2）

dminmax?=?zeros（1N）;?%?Initializes?the?vector?of?minimum?distances?
M?=?size（X1）;?%?Number?of?rows?in?X?（samples）
samples?=?1:M;

D?=?zeros（MM）;?%?Initializes?the?matrix?of?distances
for?i=1:M-1
????xa?=?X（i:）;
????for?j?=?i+1:M
??????xb?=?X（j:）;
??????D（ij）?=?norm（xa?-?xb）;
????end
end

%?D:?Upper?Triangular?Matrix
%?D（ij）?=?Euclidean?distance?between?objects?i?and?j?（j?>?i）

[maxDindex_row]?=?max（D）;?%?maxD?=?Row?vector?containing?the?largest?element?of?each?column?in?D
?????????????????????????????%?index_row（n）?=?Index?of?the?row?with?the?largest?element?in?the?n-th?column

[dummyindex_column]?=?max（maxD）;?%?index_column?=?column?corresponding?to?the?largest?element?in?matrix?D

m（1）?=?index_row（index_column）;
m（2）?=?index_column;

dminmax（2）?=?D（m（1）m（2））;

for?i?=?3:N
????%?This?routine?determines?the?distances?between?each?sample?still?available?for?selection?and?each?of?the?samples?already?selected
????pool?=?setdiff（samplesm）;?%?pool?=?Samples?still?available?for?selection
????dmin?=?zeros（1M-i+1）;?%?Initializes?the?vector?of?minimum?distances?between?each?sample?in?pool?and?the?samples?already?selected
????for?j?=?1:（M-i+1）?%?For?each?sample?xa?still?available?for?selection
????????indexa?=?pool（j）;?%?indexa?=?index?of?the?j-th?sample?in?pool?（still?available?for?selection）
????????d?=?zeros（1i-1）;?%?Initializes?the?vector?of?distances?between?the?j-th?sample?in?pool?and?the?samples?already?selected
????????for?k?=?1:（i-1）?%?The?distance?with?respect?to?each?sample?already?selected?is?analyzed
????????????indexb?=??m（k）;?%?indexb?=?index?of?the?k-th?sample?already?selected
????????????if?indexa?????????????????d（k）?=?D（indexaindexb）;
????????????else
????????????????d（k）?=?D（indexbindexa）;
????????????end
????????end
????????dmin（j）?=?min（d）;
????end
????%?The?selected?sample?corresponds?to?the?largest?dmin
????[dminmax（i）index]?=?max（dmin）;
????m（i）?=?pool（index）;
end

?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----

?????文件???????2780??2015-03-18?14:56??樣本劃分選擇\KS.m

?????文件????????543??2010-03-17?20:04??樣本劃分選擇\RS.m

?????文件???????3175??2010-03-17?20:58??樣本劃分選擇\spxy.m

?????文件?????296470??2010-03-17?20:00??樣本劃分選擇\SPXY樣本劃分法及蒙特卡羅交叉驗證結合近紅外光譜用于橘葉中橙皮苷的含量測定.PDF

?????目錄??????????0??2015-03-18?16:13??樣本劃分選擇

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???????????????302968????????????????????5